Intersection of a median of a triangle with another line segment

Question

In triangle ABC, M is the midpoint of |BC| and D is the interior point of |AB|. Point E is the intersection of the sides |AM| and |CD|. Prove that if |AD| = |DE|, then |AB| = |CE|.

I know that this is true for medians, but this should apply to all, I think.

Answer 1

If | AD | = | DE | => | AB | = | CE |.

Suppose | AD | = | DE |, denoted | AD | = x

P || AB, M∈p

ΔAED is similar ΔMFG

F∈ p∩DC

Point F is the midpoint of the segment DC, since p is a line on the middle rung ΔDBC

  FM - middle rung ΔDBC

| BD | = 2 | FM |

| BD | = 2y

| AB | = | AD | + | BD | = x + 2y

| EC | = | EF | + | FC | = y + x + y = x + 2y

So Page | AB | = | EC |

If | AD | = | DE | => | AB | = | CE |.

I got it.

Answer 2

Your question appears wrong. Suppose $ABC$ is an equilateral triangle of side 1 unit and $AD=DE=\frac12unit$. Just because $AD=DE$, it doesn't mean $AB=CE$ (as $AB=1$ and $CE=CD-DE=\frac{\sqrt5}{2}-\frac12$